Exercise 2.4.20 ($A\times B$ is divisible iff $A$ and $B$ are divisible)

Proof. Suppose that $A$ and $B$ are divisible. Let $(a,b)\in A\times B$, and let $k$ be a nonzero integer. Then there are elements $x\in A$ and $y\in B$ such that $a=x^k,b=y^k$. Therefore $(a,b)=(x^k,y^k)={(x,y)}^k$. Since this is true for every $(a,b)\in A\times B$ and every integer $k\ne 0$, where $A\times B$ is a nontrivial abelian group, $A\times B$ is a divisible group.

Conversely, suppose that $A\times B$ is divisible, where $A$ and $B$ are nontrivial abelian groups. Let $(a,b)\in A\times B$ and $k\in -\{0\}$. There is some element $(x,y)\in A\times B$ such that $(a,b)={(x,y)}^k$. Then $(a,b)=(x^k,y^k)$, so $a=x^k$ and $b=y^k$. This shows that $A$ and $B$ are divisible.

If $A$ and $B$ are nontrivial abelian groups, then $A\times B$ is divisible if and only if both $A$ and $B$ are divisible groups. □